In the series 3,9, 15,.…. What will be the \[{21^{st}}\] term?

Hint:The above question is based on the concept of geometric progression or arithmetic progression. The main approach towards solving the series is to first identify whether the sequence is arithmetic and geometric progression according to the pattern of the sequence and finding out the 21 st term according to the formula.

Complete step by step solution:
The question given has a series so we need to identify whether the series is in arithmetic progression or geometric progression. Since the difference is constant between the terms therefore the series is in arithmetic progression.
Arithmetic Progression is basically a sequence of numbers in which the difference of the consecutive numbers is a constant value.
In the above given series 3,9, 15,.….
Here the common difference is
9-3=6
15-9=6
Therefore, the common difference is 6.
So now we have to find the \[{21^{st}}\]term
So it can be written generally as to find the \[{n^{th}}\]term we have the formula
\[{a_n} = a + \left( {n - 1} \right) \times d\]
Where a is the first term in the series.
d is the common difference
n is the number of terms in the series
\[{a_n}\] is the nth term.
So since the series is in AP here a=3 and d=6,by substituting these values we get,
\[{a_{21}} = 3 + \left( {21 - 1} \right) \times 6 = 123\]
Hence the 21 st term is 123.

Note: An important thing to note is that in arithmetic progression the common difference cannot be multiplied with each term because if the common difference is in multiplication then it becomes geometric progression and for AP the common difference is always added.